Prior to reading this thread - and much to my surprise - I'd never heard of a noise bridge. I've heard of, used and actually own, a component measuring bridge, directional couplers and similar things. I've also used a noise generator. Anyway, when I meet something like this, something I know nothing about and then receive some information about it, my appetite becomes wetted for more: specifically how does it work? And why?
With this in mind, I set pencil to paper and attempted some circuit analysis. In doing this, it became apparent that it neatly dove-tails into the j Operator Thread that is running and the analysis is very akin to that applicable to a component measuring bridge: more on that item later.
The approach I have taken is to start at the basic a.c. bridge configuration and develop it into a noise bridge, step-by-step.
The basic bridge cct. is shown in figure 1.
[attachment=5200]
‘DET’ is the detector: could be any a.c.-sensing device that is suitable for the frequency and amplitude of Vs.
The conditions for balance are shown: R2 / R1 = Rx / R3. That equation is easy to prove . . . .
If the current through the ACB branch = I1; current through ACD = I2,
and voltage across R1 = Vac; across R2 = Vcb; across R3 = Vad and across Rx = Vdb,
then: Vac = I1.R1; Vcb = I1.R2; Vad = I2.R3 and Vdb = I2.Rx
If the bridge is in balance, then zero voltage appears across the detector.
Therefore, Vac = Vad and Vcb = Vdb.
So: I1.R1 = I2.R3 and I1.R2 = I2.Rx,
So: I1.R1 / I1.R2 = I2.R3 / I2.Rx,
So: R1 / R2 = R3 / Rx,
i.e.: R2 / R1 = Rx / R3.
Now we require Vs to be such that it applies a potential across points A and B.
This can be achieved by the use of a transformer with two secondary windings, suitably phased, as shown in figure 2.
[attachment=5201]
It is usual for the four windings to be quad-filar wound.
In this figure, R3 has been replaced by the variables Vc and Vr; Rx has been replaced by the unknown impedance Zx.
R1 and R2 become the secondary windings, which will have equal reactances and resistances, and thus impedances, Z1 and Z2 respectively.
Consequently the equation for balance now becomes:
Z2 / Z1 = Zx / Z3.
But Z1 = Z2, thus Zx = Z3
Figure 3 is the same arrangement as figure 2, but simply drawn in a different manner.
[attachment=5197]
However, there is a measurement limitation on this configuration for determining Zx.
If Zx is a capacitive reactance, then Zx = R – jX.
Now Z3 = R3 – jX3, so for balance:
Zx = R – jX = R3 – jX3 = Z3
Thus: R = R3 and X = X3 separately.
Hence, Zx can be determined: Zx = R3 – jX3
Note that if Zx was an inductive reactance, we would require:
R + jX = R3 – jX3, which is clearly an impossibility, since that requires X = -X3:
i.e. a capacitive reactance equal to an inductive reactance.
Hence, that circuit configuration cannot be used to determine the nature of an inductive reactance at Zx.
I found this article on Wikipedia:
http://en.wikipedia.org/wiki/Antenna_analyzer
In the Theory of Operation section, the first sentence makes no sense, neither grammatically nor technically. Nevertheless, the rest of the article, despite its brevity, is of some value.
I also found an article by ZS1JHG:
http://john-shortwavelistenersite-zs1jhg...ridge.html
In the article, his first submitted cct. is shown in figure 4.
[attachment=5198]
The reasoning for balance condition is similar to the above, but note the inclusion of C2.
The inclusion of C2 enables inductive reactance at Zx to be determined as well as capacitive reactance.
At balance:
Z1(R ± jX – jXc2) = (R1 – jXc1).Z2,
so: R ± jX – jXc2 = (R1 – jXc1).(Z2 / Z1)
Since: Z1 = Z2,
Then: R ± jX – jXc2 = R1 – jXc1.
This is only true if:
R = R1 and ± jX – jXc2 = – jXc1,
So: ± jX = j(Xc2 – jXc1),
i.e.: ±X = Xc2 – Xc1.
From that, we can see the following . . .
If: Xc2 > Xc1, X is > 0, i.e. X is inductive: Z = R + jX
If: Xc2 < Xc1, X is < 0, i.e. X is capacitive: Z = R - jX
If Xc2 = Xc1, X = 0, i.e. Z is resistive only: Z = R ± j.0
Hence, Z = = R + jX, is thus fully determined at balance:
R = R1; X = Xc2 ~ Xc1.
However, C1 and (ideally) C2 are variable-value components,
and since X is inversely proportional to C {Xc = 1/wC, numerically},
if: C2 < C1, Z is inductive;
if C2 > C1, Z is capacitive;
if C2 = C1, Z is resistive and resistive only.
The article by ZS1JHG continues with another configuration as shown in figure 5.
[attachment=5199]
Here, the ‘variables’ are a parallel arrangement, as are Z and C2.
It can be shown that that arrangement gives a result for Z that is qualitatively the same as for the series-connected variables: the bridge will handle capacitive and inductive impedances at Z. However, the resultant equations for R and X (where Z = R ± jX) are much more complicated, so that if the values of R and X need to be known, (the mathematical calculations are not trivial), as opposed to simply balancing the bridge, the series configuration is to be preferred.
(Such an arrangement is met in the Marconi TF 868B component bridge, which I will be discussing in another thread.)
And that little lot will do for now! I hope you found it useful and not too difficult to follow: I'll do my best to answer any questions arising.
Finally, there will be almost certainly some errors in the above: language / grammar and technical / mathematical. If you do spot any, please bring them to my attention.
Thank you.
Al. / June 13, 2012 //
With this in mind, I set pencil to paper and attempted some circuit analysis. In doing this, it became apparent that it neatly dove-tails into the j Operator Thread that is running and the analysis is very akin to that applicable to a component measuring bridge: more on that item later.
The approach I have taken is to start at the basic a.c. bridge configuration and develop it into a noise bridge, step-by-step.
The basic bridge cct. is shown in figure 1.
[attachment=5200]
‘DET’ is the detector: could be any a.c.-sensing device that is suitable for the frequency and amplitude of Vs.
The conditions for balance are shown: R2 / R1 = Rx / R3. That equation is easy to prove . . . .
If the current through the ACB branch = I1; current through ACD = I2,
and voltage across R1 = Vac; across R2 = Vcb; across R3 = Vad and across Rx = Vdb,
then: Vac = I1.R1; Vcb = I1.R2; Vad = I2.R3 and Vdb = I2.Rx
If the bridge is in balance, then zero voltage appears across the detector.
Therefore, Vac = Vad and Vcb = Vdb.
So: I1.R1 = I2.R3 and I1.R2 = I2.Rx,
So: I1.R1 / I1.R2 = I2.R3 / I2.Rx,
So: R1 / R2 = R3 / Rx,
i.e.: R2 / R1 = Rx / R3.
Now we require Vs to be such that it applies a potential across points A and B.
This can be achieved by the use of a transformer with two secondary windings, suitably phased, as shown in figure 2.
[attachment=5201]
It is usual for the four windings to be quad-filar wound.
In this figure, R3 has been replaced by the variables Vc and Vr; Rx has been replaced by the unknown impedance Zx.
R1 and R2 become the secondary windings, which will have equal reactances and resistances, and thus impedances, Z1 and Z2 respectively.
Consequently the equation for balance now becomes:
Z2 / Z1 = Zx / Z3.
But Z1 = Z2, thus Zx = Z3
Figure 3 is the same arrangement as figure 2, but simply drawn in a different manner.
[attachment=5197]
However, there is a measurement limitation on this configuration for determining Zx.
If Zx is a capacitive reactance, then Zx = R – jX.
Now Z3 = R3 – jX3, so for balance:
Zx = R – jX = R3 – jX3 = Z3
Thus: R = R3 and X = X3 separately.
Hence, Zx can be determined: Zx = R3 – jX3
Note that if Zx was an inductive reactance, we would require:
R + jX = R3 – jX3, which is clearly an impossibility, since that requires X = -X3:
i.e. a capacitive reactance equal to an inductive reactance.
Hence, that circuit configuration cannot be used to determine the nature of an inductive reactance at Zx.
I found this article on Wikipedia:
http://en.wikipedia.org/wiki/Antenna_analyzer
In the Theory of Operation section, the first sentence makes no sense, neither grammatically nor technically. Nevertheless, the rest of the article, despite its brevity, is of some value.
I also found an article by ZS1JHG:
http://john-shortwavelistenersite-zs1jhg...ridge.html
In the article, his first submitted cct. is shown in figure 4.
[attachment=5198]
The reasoning for balance condition is similar to the above, but note the inclusion of C2.
The inclusion of C2 enables inductive reactance at Zx to be determined as well as capacitive reactance.
At balance:
Z1(R ± jX – jXc2) = (R1 – jXc1).Z2,
so: R ± jX – jXc2 = (R1 – jXc1).(Z2 / Z1)
Since: Z1 = Z2,
Then: R ± jX – jXc2 = R1 – jXc1.
This is only true if:
R = R1 and ± jX – jXc2 = – jXc1,
So: ± jX = j(Xc2 – jXc1),
i.e.: ±X = Xc2 – Xc1.
From that, we can see the following . . .
If: Xc2 > Xc1, X is > 0, i.e. X is inductive: Z = R + jX
If: Xc2 < Xc1, X is < 0, i.e. X is capacitive: Z = R - jX
If Xc2 = Xc1, X = 0, i.e. Z is resistive only: Z = R ± j.0
Hence, Z = = R + jX, is thus fully determined at balance:
R = R1; X = Xc2 ~ Xc1.
However, C1 and (ideally) C2 are variable-value components,
and since X is inversely proportional to C {Xc = 1/wC, numerically},
if: C2 < C1, Z is inductive;
if C2 > C1, Z is capacitive;
if C2 = C1, Z is resistive and resistive only.
The article by ZS1JHG continues with another configuration as shown in figure 5.
[attachment=5199]
Here, the ‘variables’ are a parallel arrangement, as are Z and C2.
It can be shown that that arrangement gives a result for Z that is qualitatively the same as for the series-connected variables: the bridge will handle capacitive and inductive impedances at Z. However, the resultant equations for R and X (where Z = R ± jX) are much more complicated, so that if the values of R and X need to be known, (the mathematical calculations are not trivial), as opposed to simply balancing the bridge, the series configuration is to be preferred.
(Such an arrangement is met in the Marconi TF 868B component bridge, which I will be discussing in another thread.)
And that little lot will do for now! I hope you found it useful and not too difficult to follow: I'll do my best to answer any questions arising.
Finally, there will be almost certainly some errors in the above: language / grammar and technical / mathematical. If you do spot any, please bring them to my attention.
Thank you.
Al. / June 13, 2012 //
